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TrojanPoem
 one year ago
Prove that : A, B, C > 0
TrojanPoem
 one year ago
Prove that : A, B, C > 0

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434890770639:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0The value of the det is (A + B + C)^3

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I found it's value with the tri method.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0We have to prove that , The given is A , B, C > 0 so we can say their sum is also > 0 (A+B+C)^3 > 0

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But expanding this A + B +c will take up to decade .

dan815
 one year ago
Best ResponseYou've already chosen the best response.1here is a method to show this is true

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0What's the method ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1lets say ABC are all approaching the same number,

dan815
 one year ago
Best ResponseYou've already chosen the best response.1now u were to begin varying B and C,

dan815
 one year ago
Best ResponseYou've already chosen the best response.1we can rewrite B and C as Ak

dan815
 one year ago
Best ResponseYou've already chosen the best response.1and lets Say we always make sure that A is always the biggest number or equal

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0So we will make assumption that A > b,c and b, c are approaching A

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I have problem with electricity , I will relog soon , sorry.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can begin cancelling terms now

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh apparently there is a neat solution to this

dan815
 one year ago
Best ResponseYou've already chosen the best response.1heres a proof for the binomial case, maybe u can spend some time to prove it for the trinomial

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0This is the arthmetic , geometric sequence inequality right ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0The Arithmetic mean is always greater than the geometric mean

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But can you write the proof of this inequality ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1id have to work it out too,

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Are you good with maxima and minima applications and complex number ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I will open a new question
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